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3 years ago

2. What happens to the volume of a balloon when it is taken outside on a cold winter day? Why?

1 answer:

4 0

Answer: When the air particles inside the balloon become colder, they slow down and do not hit the inside of the balloon as often, so the balloon's volume decreases.

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A catalyst is used in the chemical reaction to break down hydrogen peroxide into water and oxygen. At the end of the reaction, t

Natasha_Volkova [10]

At the end of the reaction, the catalyst is UNCHANGED.
:)

8 0

3 years ago

A 30.7 g sample of Strontium nitrate, Sr(NO3)2•nH2O, is heated to a constant mass of 22.9 g. Calculate the hydration number.

Elodia [21]

Answer:

Hydration number: 4

Explanation:

1) Mass of water in the hydrated compound

Mass of water = Mass of the hydrated sample - mass of the dehydrated compound

Mass of water = 30.7 g - 22.9 g = 7.8 g

2) Number of moles of water

Number of moles = mass in grams / molar mass

molar mass of H₂O = 2×1.008 g/mol + 15.999 g*mol = 18.015 g/mol

Number of moles of H₂O = 7.9 g / 18.015 g/mol = 0.439 mol

3) Number of moles of Strontium nitrate dehydrated, Sr (NO₃)₂

The mass of strontium nitrate dehydrated is the constant mass obtained after heating = 22.9 g

Molar mass of Sr (NO₃)₂ : 211.63 g/mol (you can obtain it from a internet or calculate using the atomic masses of each element from a periodic table).

Number of moles of Sr (NO₃)₂ = 22.9 g / 211.63 g/mol = 0.108 mol

4) Ratio

0.439 mol H₂O / 0.108 mol Sr(NO₃)₂ ≈ 4 mol H₂O : 1 mol Sr (NO₃)₂

Which means that the hydration number is 4.

4 0

3 years ago

Read 2 more answers

In what state of matter do molecules bounce off one another rapidly and freely

lions [1.4K]

I think its gas i forgot

6 0

3 years ago

2.50 liter of a gas has a pressure of 165. kPa at 25.0°C. If the pressure increases to 600. kPa and the temperature to 100.0°C,

tiny-mole [99]

Answer:

0.861 L

Explanation:

We are given pressure, volume, and temperature, so let's apply the Combined Gas Law:

(P₁V₁)/T₁ = (P₂V₂)/T₂

Convert the temperatures to degrees Kelvin.

25.0°C -> 298 K, 100.0°C -> 373 K

Plug in the initial conditions on the left, then the final/new on the right, and solve for the unknown:

(165(2.5))/298 = (600(V₂))/373

V₂ = (165(2.5)(373))/(298(600))

V₂ = 0.861 L

7 0

3 years ago

Determine the acid dissociation constant for a 0.0250 M weak acid solution that has a pH of 2.37 . The equilibrium equation of i

Oksanka [162]

Answer:

For part (a): pHsol=2.22

Explanation:

I will show you how to solve part (a), so that you can use this example to solve part (b) on your own.

So, you're dealing with formic acid, HCOOH, a weak acid that does not dissociate completely in aqueous solution. This means that an equilibrium will be established between the unionized and ionized forms of the acid.

You can use an ICE table and the initial concentration ofthe acid to determine the concentrations of the conjugate base and of the hydronium ions tha are produced when the acid ionizes

HCOOH(aq]+H2O(l]⇌ HCOO−(aq] + H3O+(aq]

I 0.20 0 0

C (−x) (+x) (+x)

E (0.20−x) x x

You need to use the acid's pKa to determine its acid dissociation constant, Ka, which is equal to

3 0

2 years ago